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Anon25 [30]
3 years ago
7

A cannonball is launched at ground level at an angle of 30° above the horizontal with

Physics
2 answers:
user100 [1]3 years ago
8 0

Answer:

use the kinematics equations and solve for time, after that use dleta dx=Vi*delta time

Nookie1986 [14]3 years ago
8 0
Use the kemetic equation
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What is the velocity of a plane that traveled 3,000 miles from new york to california in 5.0
ad-work [718]
You have to divide 3000 miles with the 5 and it gives you a velocity of 600 
4 0
3 years ago
A group of hikers hear an echo 3.3 s after they shout. The temperature is 20◦C.
romanna [79]

Answer:

560 m

Explanation:

The speed of sound in air is approximately:

v ≈ v₀ + 0.6T

where v₀ is the speed of sound at 0°C (273 K) in m/s, and T is the temperature in Celsius.

The speed of sound at 20°C at that altitude is:

v ≈ 327 + 0.6(20)

v ≈ 339 m/s

The sound travels from the hikers to the mountain and back again, so it travels twice the distance.

339 m/s = 2d / 3.3 s

2d = 1118.7 m

d = 559.35 m

Rounding, the mountain is approximately 560 m away.

4 0
3 years ago
Water vapor enters a turbine operating at steady state at 500°C, 40 bar, with a velocity of 200 m/s, and expands adiabatically t
faltersainse [42]

Answer:

W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

m = 15/(2.087) = 7.17 kg/s

We also know energy equation could be defined as;

Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]

Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

4 0
3 years ago
Benny has 20 jellybeans and wants to share with his friends how many will each friend get? there are 5 friends.
trasher [3.6K]

Answer each friend will get 3.33333 repeating if he is included. if only his friends are getting them then each one gets 4

Explanation:

devide 20/6 and 20/5 respectively.

7 0
3 years ago
The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in
MissTica

Answer:

<em>155.80rad/s</em>

Explanation:

Using the equation of motion to find the angular acceleration:

\omega_f = \omega_i + \alpha t

\omega_f is the final angular velocity in rad/s

\omega_i  is the initial angular velocity in rad/s

\alpha is the angular acceleration

t is the time taken

Given the following

\omega_f = 6100rpm

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x  = 638.792rad/s

\omega_f = 638.792rad/s\\

Get the angular acceleration:

Recall that:

\omega_f = \omega_i + \alpha t

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>

5 0
2 years ago
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